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Proof 30 divides ab(a^2 -b^2)(a^2 +b^2)

  1. Sep 24, 2012 #1
    I was given the assignment
    "Show that for all integers a and b, ab(a^2 -b^2)(a^2 +b^2) is divisible by 30."

    Now, I am aware that this fails if a=b, but I want to try to prove this if we assume a≠b.

    I've reduced it to the equation (a^5)b-a(b^5), which works, and I know that I could pick the smallest two integers (1 and 2) and show that, but I don't know how to show that it exists for every integer. Should I do proof by contradiction?
     
  2. jcsd
  3. Sep 24, 2012 #2

    Mark44

    Staff: Mentor

    No, if a = b, the expression above is 0, which is divisible by 30. Note that 0/30 = 0.
    a5b - ab5 is NOT an equation.

    What do you mean by "works"? All you have done is to write the original product of three factors as the difference of two terms.

    Clearly ab(a2 - b2)(a2 + b2) = a5b - ab5, but I don't see how this is a step in the forward direction.
    For an expression to be divisible by 30, it has to be divisible by 2, 3, and 5.
     
  4. Sep 24, 2012 #3
    Yea, reducing it to [itex]a^{5}b-ab^{5}[/itex] is actually unhelpful as far as I can see, given how I proved it, as it would make things more complicated.. You want to do as Mark said and consider whether or not it is divisible by 2, 3, and 5. If it is divisible by any one of them, you don't have to worry about the factors you used affecting divisibility by the others since 2,3,5 are prime and therefore relatively prime. So just show that each of those numbers must divide the original equation and you are done.
     
    Last edited: Sep 24, 2012
  5. Sep 24, 2012 #4

    Dick

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    Homework Helper

    Right. Actually the more you factor it the easier it is. There an exhaustive way to check this if you aren't feeling particularly inspired. For example, to check if it's divisible by 5, check all pairs of values of a and b mod 5. There are 25 pairs. Using a form of symmetry will cut that in half. a=b mod 5 obviously works. Now there aren't that many pairs left. The mod 3 and mod 2 cases are much easier.
     
  6. Sep 24, 2012 #5
    The reason I had done this was not with respect to the proof, but because it made it easier for me to test whether or not this is actually divisible by 30 (my professor wants us to be in the habit of testing things and not just taking his word that they are true). Thank you for your help though.
     
  7. Sep 24, 2012 #6
    Ah I see, good 'ol testing things (but hopefully not "proving" them via American Induction, as my math prof used to joke about).
     
  8. Sep 24, 2012 #7

    Dick

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    An exhaustive test of a finite number of cases is a rigorous proof. You could have done this by doing the whole problem that way. But mod 30 you start with 900 cases. The factorization 30=2*3*5 makes it much easier.
     
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